Question

The Royal Haciendas Resort in Playa del Carmen, Mexico, is thinking of starting a new promotion. When a customer checks out of the resort after spending 5 or more days, the customer would be given a voucher that is good for 2 free nights on the next stay of 5 or more nights at the resort. The marketing manager is interested in estimating the proportion of customers who return after getting a voucher. From a simple random sample of 100 customers, 62 returned within 1 year after receiving the voucher. Please determine the 95% confidence interval estimate of the true population proportion.

Answer 1

The 95% confidence interval estimate of the true population proportion is (0.5249, 0.7151).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

From a simple random sample of 100 customers, 62 returned within 1 year after receiving the voucher.

This means that
`n = 100, \pi = (62)/(100) = 0.62`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.62 - 1.96\sqrt{(0.62*0.38)/(100)} = 0.5249`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.62 + 1.96\sqrt{(0.62*0.38)/(100)} = 0.7151`

The 95% confidence interval estimate of the true population proportion is (0.5249, 0.7151).